DM ac: Redox titrations & equations

Question 1

Explain the basic premise of manganate(VII) titrations.

Answer 1

• Acidified MnO₄^-(aq) oxidises Fe²⁺ ions to Fe³⁺ ions
• Known volume of Fe²⁺ is titrated with KMnO₄ of known concentration
• MnO₄^-(aq) is purple. No indicator is required since end point, when MnO₄^-(aq) is in excess, is when pink colour persists

Question 2

Potassium manganate(VII) oxidises iron(II) ions to iron(III) ions in acidic solution. Write the 2 half-equations and the full equation for this reaction.

Answer 2

Fe²⁺ → Fe³⁺ + e^-

MnO₄^- + 5e^- + 8H⁺ → Mn²⁺ + 4H₂O

KMnO₄^- + 5Fe²⁺ + 8H⁺ → K⁺ + 5Fe³⁺ + Mn²⁺ + 4H₂O

Question 3

The concentration of a solution of Fe²⁺ ions can be found by titration with a standard potassium manganate(VII) solution.

• Describe how the titration would be carried out.
• State how the end-point can be recognised and explain why an indicator does not need to be added.

Answer 3

• Rinse + fill burette with standard MnO₄^-(aq)
• Use graduated pipette to transfer Fe²⁺(aq) to conical flask
• Add excess H₂SO₄ to flask
• Titrate until pink colour remains due to excess MnO₄^-(aq); no indicator required
• Repeat to obtain 3 concordant results within 0.1 cm³

Don’t use HCl since it is easily oxidised to toxic Cl₂

Question 4

The end point of a manganate(VII) titration is taken as the when the pink colour of the solution in the conical flask persists for 10 seconds. Suggest why it might fade.

Answer 4

• End point was not actually reached since reaction is slow
• Reaction of MnO₄^- with water
• Reaction of MnO₄^- with Mn⁺ (forming MnO₂)

Question 5

25 cm³ of an iron(II) sulfate solution, in excess acid, requires an average titre of 17.3 cm³ 0.0200 mol dm^-3 potassium manganate(VII) solution to reach a permanent pink colour.

1. What is the concentration of Fe²⁺_(aq) in the solution?
2. The original iron(II) sulfate solution was obtained by dissolving a sample of 27.50 g in excess sulfuric acid and making it up to 2000 cm³. What was the percentage of iron in the ore? Mr of Fe = 55.8, S = 32.1, O = 16

Answer 5

5Fe²⁺_(aq) + MnO₄^-_(aq) + 8H⁺_(aq) → 5Fe³⁺_(aq) + Mn²⁺_(aq) + 4H₂O_(l)

Mol MnO₄^- = 0.02 x 17.3 x 10^-3 = 3.46 x 10^-4

Mol Fe²⁺ = 5 x 3.46 x 10^-4 = 1.73 x 10^-3

Conc Fe²⁺ in original solution = 1.73 x 10^-3 / (25 x 10^-3) = 0.069 mol dm^-3

Mol Fe²⁺ in 2000 cm³ = 0.069 x 2 = 0.138 mol

Mass FeSO₄ in 2000 cm³ = 0.138 x 151.9 = 20.96 g

% mass = 20.96 x 100 / 27.5 = 76%

Question 6

As an alternative to potassium manganate(VII) solution, potassium dichromate solution may be used in a redox titration to determine the concentration of a solution of Fe²⁺.

Explain the basic premise of this.

Answer 6

• Acidified potassium dichromate(VI) oxidises Fe²⁺ to Fe³⁺
• Known volume of Fe²⁺(aq) titrated with Cr₂O₇^2-(aq) of known concentration
• Cr₂O₇^2-(aq) is orange, so end point, when Cr₂O₇^2-(aq) is in excess, is when orange colour persists. No indicator required

Question 7

Potassium dichromate(VI) oxidises iron(II) ions to iron(III) ions in acidic solution. Write the 2 half-equations and the full equation for this reaction.

Answer 7

Cr: ox state +6 → +3, so gain of 3 electrons. But 2 Cr atoms are involved, so equation involves 6 electrons

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Fe²⁺ → Fe³⁺ + e^-

6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Question 8

Answer 8

Mol KMnO₄ = mol MnO₄^- = 0.05 x 18.4 x 10^-3 = 23/25,000

Mol C₂O₄^2- in 25 cm³ = 23/25,000 x 5/2 = 23/10,000

Mol C₂O₄^2- in 100 cm³ = 23/2500

So mol K₂C₂O₄ • H₂O = 23/2500

Mr K₂C₂O₄ • H₂O = 184.2

Mass K₂C₂O₄ • H₂O = 23/2500 x 184.2 = 1.69 g

Question 9

Answer 9

Mol C₂O₄^2- = 0.05 x 10 x 10^-3 = 1/2000

Mol MnO₄^- = 1/2000 x 2/5 = 1/5000

Conc MnO₄^2- = 1/5000 / (26 x 10^-3) = 0.00769 mol dm^-3

Question 10

Answer 10

Mol S₂O₃^2- = 0.2 x 20.5 x 10^-3 = 41/10,000

Mol Cu²⁺ in 25 cm³ = 41/10,000

Mol Cu²⁺ in 250 cm³ = 41/10,000 x 250/25 = 41/1000

Mass Cu²⁺ in coin = 41/1000 x 63.5 = 2.6035 g

% mass Cu = 2.6035 x 100 / 3.47 = 75.0 %

Question 11

When zinc is added to copper(II) sulfate solution, it displaces the copper.

Suggest two changes which would be observed.

Answer 11

• Blue solution becomes paler
• Copper metal deposits on zinc

Question 12

Balance the following equation:

Sn + HNO₃ → SnO₂ + NO₂ + H₂O

Answer 12

Sn + HNO₃ → SnO₂ + NO₂ + H₂O

Balance ox states:

Sn: 0 → 4 (gains 4 e^-)

N: 5 → 4 (loses 1 e^-)

Sn + 4HNO₃ → SnO₂ + 4NO₂ + H₂O

No charge to balance

Balance rest

O: 12 → 11

H: 4 → 2

Sn + 4HNO₃ → SnO₂ + 4NO₂ + 2H₂O

Question 13

Balance the following equation:

MnO₄^- + Fe²⁺ + H⁺→ Mn²⁺ + Fe³⁺ + H₂O

Answer 13

MnO₄^- + Fe²⁺ + H⁺→ Mn²⁺ + Fe³⁺ + H₂​O

Balance ox states:

Mn: +7 → +2 (gains 5 e^-)

Fe: +2→ +3 (loses 1 e^-)

MnO₄^- + 5Fe²⁺ + H⁺→ Mn²⁺ + 5Fe³⁺ + H₂​O

Balance charge

As is: +10 → +17

Only alter species not involved in redox:

MnO₄^- + 5Fe²⁺ + 8H⁺→ Mn²⁺ + 5Fe³⁺ + 4H₂​O

No rest to balance

Question 14

Write a half equation to represent the change when an aqueous solution of chlorate(I), ClO^-, ions is reduced, forming chloride ions and hydroxide ions.

Answer 14

ClO^- + 2e^- + H₂O → Cl^- + 2OH^-

Question 15

Hydrogen peroxide solution was added to a test tube containing sodium chlorate(I) solution.

• Decide which species are reduced and oxidised and deduce the two half-equations
• Write the redox requation
• Use this to suggest what is observed

Answer 15

Reduction:

ClO^- + 2e^- + H₂O → Cl^- + 2OH^-

+1 to -1: gain of 2

Oxidation:

H₂O₂ → O₂ + 2H⁺ + 2e^-

-1 to 0: loss of 1

Overall:

ClO^- + H₂O + H₂O₂ → Cl^- + 2OH^- + O₂ + 2H⁺

Observation: Fizzing as oxygen gas forms

Question 16

1. Write the balanced redox equation for the reaction between hydrogen peroxide and iodide ions in acidic solution.
2. Use this to suggest what is observed.

Answer 16

Reduction:

H₂O₂ + 2H⁺ + 2e^- → 2H₂O

-1 to -2: gain of 1

Oxidation:

2I^- → I₂ + 2e^-

-1 to 0: loss of 1

Redox:

H₂O₂ + 2H⁺ + 2I^- → 2H₂O + I₂

Observation: Colourless to brown as I₂ forms

Question 17

A student carried out the following steps:

• Add copper(II) sulfate solution to a test tube containing potassium iodide solution
• Add sodium thiosulfate solution until no further change

1. Write the balanced redox equation for the reaction before thiosulfate is added
2. Describe what happens next, using equations to support your answer
3. Give the overall reaction equation
4. Use this to suggest what is observed

Answer 17

1. Reduction: Cu²⁺ + e^- → Cu⁺

+2 to +1: gain of 1

Oxidation: 2I^- → I₂ + 2e^-

-1 to 0: loss of 1

Redox: 2Cu²⁺ + 2I^- → I₂ + 2Cu⁺

2. Thiosulfate reacts with iodine, reforming iodide ions, which then react with Cu(I) ions:

I₂ + 2S₂O₃^2- → 2I^- + S₄O₆^2-

2Cu⁺ + 2I^- → 2CuI

3. 2Cu²⁺ + 2I^- + 2S₂O₃^2- → 2CuI + S₄O₆^2-

4. Colourless to brown as I₂ forms, then, on addition of thiosulfate, white CuI precipitate forms

Question 18

Balance the following equation:

Cr₂O₇^2- + H⁺ + I^- → I₂ + Cr³⁺ + H₂O

Answer 18

Cr₂O₇^2- + H⁺ + I^- → I₂ + Cr³⁺ + H₂​O

Balance ox states:

Cr: +6 → +3 (gain of 3)

I: -1 → 0 (loss of 1)

For every Cr atom, there are 3 I atoms

Cr₂O₇^2- + H⁺ + 6I^- → 3I₂ + Cr³⁺ + H₂​O

Balance charge / rest:

Cr₂O₇^2- + 14H⁺ + 6I^- → 3I₂ + Cr³⁺ + 7H₂O​

Question 19

Balance the following equation:

H⁺ + H₂O₂ + Fe²⁺ → H₂O + Fe³⁺

Answer 19

H⁺ + H₂O₂ + Fe²⁺ → H₂O + Fe³⁺

Balance ox states:

O: -1 to -2 (gain of 1)

Fe: +2 to +3 (loss of 1)

For every O atom, there is 1 Fe atom

H⁺ + H₂O₂ + 2Fe²⁺ → 2H₂O + 2Fe³⁺

Balance charge / rest:

2H⁺ + H₂O₂ + 2Fe²⁺ → 2H₂O + 2Fe³

Question 20

When chlorine reacts with hot hydroxide ions, part of it is oxidised to chlorate(V) and the rest is reduced to chloride. Balance the equation for this reaction:

Cl₂ + OH^- → ClO₃^- + Cl^- + H₂O

Answer 20

Cl₂ + OH^- → ClO₃^- + Cl^- + H₂O

Balance ox states:

Nothing but chlorine is oxidised or reduced. So Cl₂ acts as both agents.

0 to +5 / -1 (gain/loss of 0)

3Cl₂ + OH^- → ClO₃^- + 5Cl^- + H₂O

Balance charge:

3Cl₂ + 6OH^- → ClO₃^- + 5Cl^- + H₂O

Balance rest:

3Cl₂ + 6OH^- → ClO₃^- + 5Cl^- + 3H₂O

Question 21

Answer 21

C

Question 22

The ‘kappa number’ of wood pulp is used to measure the amount of oxidisable material the pulp contains. The kappa number is defined as the volume (cm³) of acidified 0.020 mol dm^-3 KMnO₄ that will react with 1.0 g of dry wood pulp.

The pulp is weighed, then broken up in water, then 25.0 cm³ of a solution containing 0.020 mol dm^-3 MnO₄^- is added. The mixture is stirred for 5 minutes at 25 °C.

Excess iodide ions are added. The remaining MnO₄^- ions react with the iodide ions to form iodine. The iodine formed is titrated with a solution containing S₂O₃^2- ions:

2S₂O₃^2- + I₂ → S₄O₆^2- + 2I^-

a) Complete the equation below for the reaction of acidified MnO₄^- ions with iodide.

2MnO₄^- + ……………… + …………… → 5I₂ + ……………… + ……………… (2 marks)

A 2.5 g sample of dry wood pulp is treated as described and requires 7.5 cm³ of a solution containing 0.20 mol dm^-3 S₂O₃^2- ions.

b) Calculate the kappa number of the wood pulp. (6 marks)

Answer 22

a) 2MnO₄^- + 10I^- + 16H⁺ → 5I₂ + 2Mn²⁺ + 8H₂O

b) Mol S₂O₃^2- = 0.2 x 7.5 x 10^-3 = 0.0015 mol

Mol I₂ 0.0075 mol

Mol unreacted MnO₄^- = 0.0003 mol

Mol MnO₄^- added = 0.02 x 25 x 10^-3 = 0.0005 mol

Mol reacted MnO₄^- = 0.0005 - 0.0003 = 0.0002 mol

Vol MnO₄^- = 0.0002/0.02 = 0.01 dm³ = 10 cm³

10 cm³ for 2.5 g → 4 cm³ for 1 g